How to show that such a function is continuous

Question

Let $G$ be a compact group with a fixed Haar measure $dy$. Let $F_1$ be a complex-valued continuous function on $G$. Set $F_2(x):=\int_G F_1(yxy^{-1})dy$ for $x\in G$. Show that $F_2$ is continuous.

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Could anyone give a hint? Thanks!

Answer 1

Define the group $H = G \times G$ and the map $L : H \to \mathbb R$ by $L(x,y)= F_1(yxy^{-1})$

Cover $\mathbb R$ by balls of radius $\epsilon$. Each ball has an open preimage under $L$. Let $\mathcal U$ be the cover of $H$ by there preimages. By the Lebesgue number lemma there is an open neighborhood $U \subset H$ of the identity whose every translate is contained entirely in an element of $\mathcal U$. That means:

$(x,y)(a,b)^{-1} \in U \implies |F_1(yxy^{-1}) - F_1(bab^{-1})| < 2\epsilon$

That means there are neighborhoods $V$ and $W$ of the identity in $G$ such that. . .

$(xa^{-1},yb^{-1}) \in V \times W \implies |F_1(yxy^{-1}) - F_1(bab^{-1})| < 2\epsilon$

In particular when $y=b$ we have. . .

$xa^{-1} \in V \implies |F_1(yxy^{-1}) - F_1(yay^{-1})| < 2\epsilon$

Observe the LHS is independent of $y$.

Now integrate the RHS over $G$ to recover the definition of $F_2$ being continuous at $a \in G$.

Answer 2

Suppose $x_n \to x$. Then $F_2(x_n) = \int_G F_1(yx_ny^{-1})dy$. Since $F_1$ is continuous and $G$ is compact, say $|F_1(z)| \le C$ for all $z \in G$. The point is that since $C \in L^1(G)$ (since Haar measure on compact group is finite), we can use dominated convergence, noting $F_1(yx_ny^{-1}) \to F_1(yxy^{-1})$ for each $y$, by the continuity of $F_1$ and the group operations. So, $F_2(x_n) \to \int_G F_1(yxy^{-1})dy = F_2(x)$.

Disclaimer: I might have cheated a bit. It might not be true that "if $x_n \to x$, then $F_2(x_n) \to F_2(x)$" implies $F_2$ is continuous. We might need to use nets...